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M23-20

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M23-20  [#permalink]

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New post 16 Sep 2014, 00:19
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Difficulty:

  25% (medium)

Question Stats:

80% (01:55) correct 20% (02:21) wrong based on 160 sessions

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If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

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Re M23-20  [#permalink]

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New post 16 Sep 2014, 00:19
Official Solution:

If a car had traveled 20 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 60 km, at what speed did it travel?

A. 35 kmh
B. 40 kmh
C. 50 kmh
D. 60 kmh
E. 65 kmh

Solve the equation \(\frac{60}{x} = \frac{60}{x + 20} + 0.5\). Backsolve: \(\frac{60}{40} = \frac{60}{40 + 20} + 0.5\).

Answer: B
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M23-20  [#permalink]

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New post 19 Dec 2016, 09:08
Hi,

Is it possible to calculate it algebraically ?

I get stuck with two unknown, even if I know that is quicker to back solve, I do not understand why it is not solvable ?

Many thanks in advance for your kind help
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Re: M23-20  [#permalink]

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New post 19 Dec 2016, 09:18
nickimonckom wrote:
Hi,

Is it possible to calculate it algebraically ?

I get stuck with two unknown, even if I know that is quicker to back solve, I do not understand why it is not solvable ?

Many thanks in advance for your kind help


Which two unknowns are you talking about? There is only one unknown in the equation above. If you simplify \(\frac{60}{x} = \frac{60}{x + 20} + 0.5\) you'll get quadratic equation which will give two values of x: -60 (discard) and 40.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M23-20  [#permalink]

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New post 20 Dec 2016, 01:53
I mean that if I set up the table, I get the following :

FIRST MOVE : \(r*(t+\frac{1}{2}) = 60\)
2ND MOVE : \((r+20)*t = 60\)

Then, since they are equal :
\(r*(t+\frac{1}{2}) = (r+20)*t\)
\(rt + \frac{r}{2} = rt + 20t\)
\(r = 40t\)

Then I have a problem with the unit, I know that 40 is the good value but I am not supposed to express a rate and a time, only a rate.

thanks in advance for your help
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Re: M23-20  [#permalink]

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New post 06 Apr 2017, 21:33
Could you please share some links to questions similar to this? Thank you.
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Re: M23-20  [#permalink]

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New post 06 Apr 2017, 23:29
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Re: M23-20  [#permalink]

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New post 09 Apr 2018, 11:05
For those of you wanting to set up an equation and solve it w/o plugging in, I can chime in to help.
Since traveling at 20kmh faster saves 1/2 hour, we can say that:
60/R - 60/(R+20) = 1/2, which is basically saying:
(Time at Reg Speed) - (Time +20kmh faster) = (30 mins)

Solving, we get:
(R+60)(R-40)=0
R= -60, 40

Rate can not be negative, therefore, 40 is the winner.
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Re: M23-20 &nbs [#permalink] 09 Apr 2018, 11:05
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